Question

CaCl₂(aq) + 2 AgNO₃(aq) -->Ca(NO₃)₂(aq) + 2 AgCl(s)

What mass of silver chloride will precipitate when 30.0 mL of 0.50 mol/L calcium chloride, CaCl₂(aq), react completely with silver nitrate solution?

Make sure you have the correct number of significant digits and units. 

a) Calculate the moles of calcium chloride.

c) Calculate the mass of silver chloride.

Answer 1

Solution.

$V(CaCl_2)=30.0mL=0.03L;$

$c(CaCl_2)=0.50mol/L;$

$a)c=\dfrac{\nu}{V}\implies \nu=c\sdot V;$

$\nu(CaCl_2)=0.50mol/L\sdot0.03L=0.015mol;$

$c)$ $0.015mol$ $x mol$

$CaCl_2(aq) + 2 AgNO_3(aq) -->Ca(NO_3)_2(aq) + 2 AgCl(s);$

1 2

$x=\dfrac{0.015mol\sdot 2}{1}=0.03mol;$

$\nu(AgCl)=0.03mol;$

$M(AgCl)=143.32 g/mol;$

$\nu=\dfrac{m}{M}\implies m=\nu\sdot M;$

$m(AgCl)=0.03mol\sdot143.32g/mol=4.30g;$

Answer: $a)$ $\nu(CaCl_2)=0.015mol;$

$c) m(AgCl)=4.3g.$